NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

Question 1: For a positive integer n, find the value of $(1 - i)^n$ $(1-\frac 1 i)^n$

Answer:

$(1-i)^n (1-1/i)^n$
$=(1-i)^n (1+i )^n$
$=(1-i^2 )^n$
$=2^n$

Question 2 Evaluate $\sum_{i=1}^{13}$$(i^n+i^{n+1} )$ where n $\epsilon$N.

Answer:

$\sum_{i=1}^{13}$$(i^n+i^{n+1} )$
$=$ $\sum_{i=1}^{13}$$(1+i) i^n$
$=(1+i)(1+i^2+i^3+i^4....+i^{12}+i^{13} )$
$=(1+i) \frac{i(i^{13}-1)}{(i-1)}$
$=(1+i) i\frac{(i-1)}{(i-1)}$
$=(1+i)i$
$=i+i^2$
$=i-1$

Question 3 If $\frac{(1+i)}{(1-i)}^3-\frac{(1-i)}{(1+i)}^3= x+iy$, then find (x, y).

Answer:

Given $\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3= x+iy$
$=\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3$
$=\left (\frac{1+2i+i^2}{1-i^2} \right )^{3}-\left (\frac{1-2i+i^2}{1-i^2} \right )^{3}$
$=(\frac{2i}{2})^3-(\frac{-2i}{2})^3$
$=i^3-(-i^3 )$
$=2i^3=0-2i$
$Thus,(x,y)=(0,-2)$

Question 4 If $(1+i)^2/(2-i)= x+iy$ then find the value of x + y.

Answer:

$x+iy=\frac{(1+i)^2}{(2-i)}=\frac{ (1+2i+i^2)}{(2-i)}= \frac{2i}{(2-i)}$

$On\: \: rationalising\: \: 2i(2+i)/(2-i)(2+i)$,
$=\frac{(4i+2i^2)}{(4-i^2 )}=\left (\frac{(4i-2)}{(4+1)} \right )$
$=\frac{-2}{5}+\frac{4i}{5}$
$x=\frac{-2}{5} \: \: and\: \: y=\frac{4}{5}$

$So, x+y= -\frac{2}{5}+\frac{4}{5}=\frac{2}{5}$

Question 5 If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$ then find (a, b).

Answer:

$a+ib=\left(\frac{1-i}{1+i}\right)^{100}=\left [\frac{1-i}{1+i}*\frac{1-i}{1-i} \right ]^{100}$
$=\left [\frac{(1-i)^2}{1-i^2} \right ]^{100}$
$=\left (\frac{(1-2i+i^2)}{(1+1)} \right )^{100}$
$=\left (\frac {-2i}{2 }\right )^{100}$
$=(i^4 )^{25}=1$
$Hence,(a,b)=(1,0)$

Question 6 If $a=\cos \theta+i \sin\theta$, find the value of$\frac{(1+a)}{(1-a)}$ .

Answer:

$a=\cos \theta+i \sin\theta$
$\frac{(1+a)}{(1-a)}$$=\frac{(1+cos\theta)+i \sin \theta}{(1-cos\theta)-isin\theta}$

$=\frac{\left ( 2\cos ^{2}\frac{\theta }{2} +i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}{\left ( 2\sin ^{2}\frac{\theta }{2} -i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}$
$=\frac{2\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{2\sin\frac{\theta }{2}\left ( \sin\frac{\theta }{2}-i\cos\frac{\theta }{2} \right )}$
$=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}-i^2\cos\frac{\theta }{2} \right )}$
$=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}+\cos\frac{\theta }{2} \right )}$
$=i\cot \frac{\theta }{2}$

Question 7 If $(1 + i)z = (1 - i) \bar{z},$ then show that $z= -i\bar{z}$

Answer:

$(1 + i)z = (1 - i) \bar{z},$
$z=\frac{1-i}{1+i}\bar{z}$
Rationalising the denominator,
$\frac{\left (1-i \right )\left (1-i \right )}{\left (1+i \right )\left (1-i \right )}\bar{z}$
$=\frac{\left ( 1-i \right )^{2}}{1-i^2}\bar{z}$
$=\frac{\left ( 1-2i+i^2 \right )}{1+1}\bar{z}$
$=\frac{\left ( 1-2i-1 \right )}{2}\bar{z}$
$=-i\bar{z}$
Hence, Proved.

Question 8: If$z = x + iy$, then show that $z\bar{z}+2(z+\bar{z})+b=0$ where b belongs to R, representing z in the complex plane as a circle.

Answer:

$z=x+iy$
$\bar{z}=x-iy$
$Now,we\: \: also\: \: have\: \: ,z\bar{z }+2(z+\bar{z})+b=0$
$(x+iy)(x-iy)+2(x+iy+x-iy)+b=0$
This equation is an equation of a circle.

Question 9: If the real part of $\frac{(\bar{z}+2)}{(\bar{z}-1)}$ is 4, then show that the locus of the point representing z in the complex plane is a circle.

Answer:

$Let \: \: z=x+iy \: \:$
$\frac{\bar{z}+2}{\bar{z}-1}= \frac{x-iy+2}{x-iy-1}$
$=\frac{\left [ \left ( x+2 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}{\left [ \left ( x-1 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}$
$=\frac{(x-1)(x+2)+y^2+i[(x+2)y-(x-1)y]}{(x-1)^2+y^2}$
$real \: \: part=4 \Rightarrow \frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4$
$x^2+x-2+y^2=4(x^2-2x+1+y^2 )$
$3x^2+3y^2-9x+6=0$

The equation obtained is that of a circle. Hence, the locus of z is a circle.

Question 10 Show that the complex number z, satisfying the condition $arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}$ lies on a circle.

Answer:

Let z=x+iy
$arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}$
$arg\left ( z-1 \right )-arg\left ( z+1 \right )=$$\frac{\pi}{4}$
$arg\left ( x+iy-1 \right )-arg\left ( x+iy+1 \right )=$$\frac{\pi}{4}$
$arg\left ( x-1+iy \right )-arg\left ( x+1+iy \right )=$$\frac{\pi}{4}$
$\tan^{-1}\frac{y}{x-1}-\tan^{-1}\frac{y}{x+1}=$$\frac{\pi}{4}$
$\tan^{-1}\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left (\frac{y}{x-1} \right )\left (\frac{y}{x+1} \right )}=$$\frac{\pi}{4}$
$\tan^{-1}\left(\frac{y\left ( x+1+-x+1 \right )}{x^2-1+y^2}\right) = \frac{\pi}{4}$
$\tan\frac{\pi}{4} = \frac{2y}{x^2+y^2-1}$
$x^2+y^2-1=2y$
$x^2+y^2-1-2y=0$
The equation obtained represents the equation of a circle

Question 11 Solve that equation $|z| = z + 1 + 2i$.

Answer:

$|z| = z + 1 + 2i$
$Substituting \: \: z=x+iy \: \: we \: \: get \: \: |x+iy|=x+iy+1+2i$
$|z|= \sqrt{(x^2+y^2 )} =(x+1)+i(y+2)$
Comparing real and imaginary parts $\sqrt{(x^2+y^2 )} =(x+1)$
And$0=y+2 , y=-2$
Substituting the value of y in$\sqrt{(x^2+y^2 )} =(x+1)$
$x^2+(-2)^2=(x+1)^2$
$x^2+4=x^2+2x+1$

Hence, $x=\frac{3}{2}$
Hence,$z=x+iy$
$=\frac{3}{2} -2i$

Question 12 If $|z + 1| = z + 2 (1 + i)$, then find z

Answer:

We have $|z + 1| = z + 2 (1 + i)$
Substituting $z=x+iy$ we get $x+iy+1=x+iy+2i+1$
$|z|= \sqrt{(x^2+y^2 ) }=\sqrt{(x+1)^2+y^2 } = (x+2)+i(y+2)$
Comparing real and imaginary parts, $\sqrt{(x+1)^2+y^2 } = (x+2)$
$and 0=y+2 ;y=-2$
Substituting the value of y in $\sqrt{(x+1)^2+y^2 } = (x+2)$
$(x+1)^2+(-2)^2=(x+2)^2$
$x^2+2x+1+4=x^2+4x+4$
$2x=1$
Hence, $x=\frac{1}{2}$
Thus, $z=x+iy=\frac{1}{2}-2i$

Question 13 If $arg (z - 1) = arg (z + 3i)$, then find x – 1 : y. where $z = x + iy$

Answer:

Given that $arg (z - 1) = arg (z + 3i)$
$arg (x+iy-1)=arg(x+iy+3i)$
$arg(x-1+iy)=arg(x+i(y+3))$
$\tan^{-1}\frac{y}{x-1}=\tan^{-1}\frac{y+3}{x}$
$\frac{y}{x-1}=\frac{y+3}{x}$
$xy=xy-y+3x-3$
$3x-3=y$
$\frac{x-1}{y}=\frac{1}{3}$

Question 14 Show that $\left | \frac{z-2}{z-3} \right |=2$ represents a circle. Find its centre and radius.

Answer:

$\left | \frac{z-2}{z-3} \right |=2$ Substituting $z=x+iy$, we get $\left | \frac{x+iy-2}{x+iy-3} \right |=2$
$|x-2+iy|=2|x-3+iy|$
$\sqrt{(x-2)^2+y^2 }=2\sqrt{((x-3)^2+y^2}$
$x^2-4x+4+y^2=4(x^2-6x+9+y^2 )$
$3x^2+3y^2-20x+32=0$
$x^2+y^2-\frac{20}{3} x+\frac{32}{}3=0$
$(x-\frac{10}{3})^2+y^2+\frac{32}{3}-\frac{100}{9}=0$
Thus, the centre of circle is $(\frac{10}{3},0)$ and radius is $2/3$

Question 15 If $\frac{z-1}{z+1}$ is a purely imaginary number$(z \neq -1)$, then find the value of $|z|$.

Answer:

Let $z=x+iy$
$\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$
$\frac{\left [ \left ( x-1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}{\left [ \left ( x+1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}$
$=\frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2 }$
$\frac{z-1}{z-1}$ is purely imaginary
$\frac{(x-1)(x+1)+y^2}{(x+1)^2+y^2 }=0$
$x^2-1+y^2=0$
$x^2+y^2=1$ Hence, $|z|=1$

Question 16 z1 and z2 are two complex numbers such that $|z_1| = |z_2|$ and $arg (z_1) + arg (z_2) = \pi$, then show that $|z_1| = -|z_2|$ .

Answer:

Let $z_1=\left |z_1 \right |\left (cos\theta _1+i\ sin\theta_1 \right )$ and $z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$
Given that $|z_1| = |z_2|$
And $arg\left (z_1 \right )+arg\left (z_2 \right )= \pi$

$\theta_1+\theta_2=\pi$

$\theta_1=\pi-\theta_2$

$z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$

$z_1=|z_2 |\left ( -\cos \theta_2+isin\theta_2 \right )$

$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$

$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$, $|z_1| = -|z_2|$

Question 17 If |z1| = 1 (z1 ≠ –1) and $z_2= \frac{z_1-1}{z_1+1}$ then show that the real part of z2 is zero

Answer:

Let $z_1=x+iy$ $|z_1 |= \sqrt{x^2+y^2} =1$
$z_2= \frac{z_1-1}{z_1+1}=\frac{x+iy-1}{x+iy+1}$
$=\frac{[(x-1)+iy][(x+1)-iy]}{[(x+1)+iy][(x+1)-iy] }$
$= \frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2}$
$\frac{x^2+y^2-1+2iy}{(x+1)^2+y^2 }=0$
Since, $x^2+y^2=1$
$\frac{1-1+2iy}{(x+1)^2+y^2}=\frac{0+2iy}{(x+1)^2+y^2}$
Therefore, the real part of $z_2$ is zero

Question 18 If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then find $arg\left ( \frac{z_1}{z_4} \right )+arg\left ( \frac{z_2}{z_3} \right )$

Answer:

$z_1$ and $z_2$ are conjugate complex numbers.
The negative side of the real axis $= r_1 (cos\theta-isin \theta)$
$=r_1\left ( \cos\left ( -\theta_1 \right )+i\sin\left (- \theta_1 \right ) \right )$
Similarly, $z_3=r_2\left ( \cos\left ( \theta_2 \right )-i\sin\left ( \theta_2 \right ) \right )$
$z_4=r_2\left ( \cos\left ( -\theta_2 \right )+i\sin\left (- \theta_2 \right ) \right )$
$arg \left ( \frac{z_1}{z_4} \right )+arg \left ( \frac{z_2}{z_3} \right )$$=arg(z_1)-arg(z_4)+ arg(z_2)- arg(z_3)??$
$=\theta_1-(-\theta_2)+(-\theta_1)-\theta_2=0$

Question 19 If $|z_1| = |z_2| =... = |z_n| = 1$, then
Show that $|z_1 + z_2 + z_3 +... + z_n| =|1/z_1 +1/z_2 +...+1/z_n |$

Answer:

$|z_1| = |z_2| =... = |z_n| = 1$
$|z_1|^2 = |z_2|^2 =... = |z_n|^2 = 1$
$z_1 \bar{z_1 } =z_2 \bar{z_2 }=...=z_n \bar{z_n }$
$z_1=\frac{1}{\bar{z_1}},z_2=\frac{1}{\bar{z_2}},...z_n=\frac{1}{\bar{z_n}}$
Now, $|z_1+z_2+z_3+...z_n |$ $=\left | \frac{z_1\bar{z_1}}{\bar{z_1}}+\frac{z_2\bar{z_2}}{\bar{z_2}}+...+\frac{z_n\bar{z_n}}{\bar{z_n}} \right |$
=$\left | \frac{1}{\bar{z_1}}+\frac{1}{\bar{z_2}}+...+\frac{1}{\bar{z_n}} \right |$
$=\left | {\frac{1}{z_1}+\frac{1}{z_2}+...+\frac{1}{z_n}} \right |$

Question 20 If for complex numbers z1 and z2, $arg (z_1) - arg (z_2) = 0$, then show that $\left | z_1-z_2 \right |= |z_1|-|z_2|$

Answer:

Let $z_1=\left | z_1 \right |\left ( \cos\theta_1 +i\sin \theta_1 \right )$ and $z_2=\left | z_2 \right |\left ( \cos\theta_2 +i\sin \theta_2 \right )$
$arg(z_1 )- arg(z_1 )=0$
$\theta _1-\theta _2=0$
$\theta _1=\theta _2$
$z_1-z_2= (z_1 |cos\theta_1-| z_2|cos\theta_1 )+i(z_1 |sin\theta_1-| z_2|sin\theta_1 )$
$z_1-z_2= \sqrt{(z_1 |cos\theta_1-| z_2|cos\theta_1 )^{2}+(z_1 |sin\theta_1-| z_2|sin\theta_1 )^{2}}$
$=\sqrt{|z_1|^2+|z_2|^2-2|z_1||z_2|\cos^2\theta_1-2|z_1||z_2|\sin^2\theta_1}$
$=\sqrt{|z_1|^2+|z_2|^2-2|z_1||z_2|}$
$=\sqrt{\left (|z_1|-|z_2| \right )^2}= |z_1|-|z_2|$

Question 21 Solve the system of equations $Re (z^2) = 0, |z| = 2.$

Answer:

$Re (z^2) = 0, |z| = 2.$
$Let\: \: z=x+iy$
$|z|= \sqrt{x^2+y^2 } =2$
${x^2+y^2 } =4$
$z^2=x^2+2ixy-y^2$
$=(x^2-y^2 )+2ixy$
$Now,Re (z^2 )=0$
$x^2-y^2=0$
$x^2=y^2=2$
$x=y=\pm \sqrt{2}$
Hence, $z=x+iy=\pm\sqrt{2}\pm i\sqrt{2}$
$=\sqrt{2}+i\sqrt{2},\sqrt{2}-i\sqrt{2},-\sqrt{2}+i\sqrt{2} \: \: and\: \: -\sqrt{2}-i\sqrt{2}$

Question 22 Find the complex number satisfying the equation $z+\sqrt{2} |(z+1)|+i=0$.

Answer:

$z+\sqrt{2} |(z+1)|+i=0\: \: \: \: ...(i)$
Substituting $z=x+iy$ we get $x+iy+\sqrt2 |x+iy+1|+i=0$
$x+i(1+y)+\sqrt2 \left [ \sqrt{(x+1)^2+y^2 } \right ]=0$
$x+i(1+y)+\sqrt2 \left [ \sqrt{(x^2+2x+1+y^2 ) } \right ]=0$
1+y = 0
y = -1
$x+\sqrt2 \sqrt{x^2+2x+2}=0$
$\sqrt2 \sqrt{x^2+2x+2}=-x$
$2x^2+4x+4=x^2$
$x^2+4x+4=0$
$(x+2)^2=0$
$x=-2$
Hence,$z=x+iy=-2-i$

Question 23 Write the complex number $z=\frac{1-i}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$ in polar form

Answer:

$z=\frac{1-i}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\frac{\sqrt{2}\left [ \frac{1}{\sqrt{2}}-i1/\sqrt{2} \right ]}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\frac{\sqrt{2}\left [\cos\frac{\pi}{4}-isin\frac{\pi}{4} \right ]}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\sqrt2\left [ \cos \left ( -\frac{\pi}{4}-\frac{\pi}{3}\right )+i\sin\left (-\frac{\pi}{4}-\frac{\pi}{3} \right ) \right ]$
$=\sqrt2\left [ \cos \left ( -\frac{7\pi}{12}\right )+i\sin\left (-\frac{7\pi}{12} \right ) \right ]$
$=-\sqrt2\left [ \cos \left ( \frac{5\pi}{12}\right )+i\sin\left (\frac{5\pi}{12} \right ) \right ]$

Question 24 If z and w are two complex numbers such that $|zw| = 1$ and $arg(z) - arg (w) = \pi/2$, then show that $zw=-i$.

Answer:

Let $z=\left | z \right |(\cos\theta_1 + i sin\theta_1 )$ and $w=\left | w \right |(\cos\theta_2 + i sin\theta_2 )$
$|zw|=|z||w|=1$
Also $arg(z)-arg(w)= \frac{\pi}{2}$
$\theta_1-\theta_2=\frac{\pi}{2}$
$\bar{z} w=|z|\left (\cos \theta _{1}-i\sin\theta _{1} \right )w$
$w=|w|\left (\cos \theta _{2}+i\sin\theta _{2} \right )=1$
$\bar{z}w$$=|z||w|(\cos (-\theta_1)+i sin(-\theta_1 ))(cos\theta_2+i \sin\theta_2 )$
$\bar{z}w$$=\cos [(\theta _2 - \theta _1 )+isin( \theta _2- \theta _1 )]$
$=\left [ \cos \left ( -\frac{\pi}{2} \right ) +i\sin\left ( -\frac{\pi}{2} \right )\right ]$
$=1[0-i]=-i$

Question 25 i) For any two complex numbers z1, z2 and any real numbers a, b, $|az_1 - bz_2|^2 + |bz_1 + az_2|^2 =....$ ii) The value of $\sqrt{-25} * \sqrt{-9}$ is iii) The number $\frac{(1-i)^3}{1-i^3}$ is equal to ....... iv) The sum of the series $i+i^2+i^3+...+i^{1000}$ upto 1000 terms is ... v) Multiplicative inverse of 1 + i is ................ vi) If $z_1 \ and \ z_2$ are complex numbers such that $z_1 + z_2$ is a real number, then $z_2$ = vii) $arg (z)+arg (\bar{z} )$ $(\bar{z}\neq0)$ is .... viii) If $|z+4|\leq 3$then the greatest and least values of |z+1| are ..... and ..... ix) if $\frac{z-2}{z+2}=\frac{\pi}{6}$then the locus of z is ....... x) If $|z|=4$ and $arg (z) = \frac{5\pi}{6}$ then z=

Answer:

i) $|az_1 - bz_2|^2 + |bz_1 + az_2|^2$
$=|az_1 |^2+|bz_2 |^2-2Re(az_1.b(\bar{z_2} ) +|bz_1 |^2+|az_2 |^2+2Re(az_1.b\bar{z_2} ) ?)$
$=|az_1 |^2+|bz_2 |^2+|bz_1 |^2+|az_2 |^2=(a^2+b^2 )(|z_1 |^2+|z_2 |^2 )$
ii)$\sqrt{-25}*\sqrt{-9}=5i*3i=15i^2=-15$
iii) $\frac{(1-i)^3}{1-i^3}=\frac{(1-i)^3}{(1-i)(1+i+i^2 )}$
$=\frac{(1-i)^3}{(1+i-1) }= \frac{1+i^2-2i}{}i$
$=\frac{1-1-2i}{i}=-\frac{2i}{i}=-2$
iv) $i+i^2+i^3+...+i^{1000}=0 \left [ \sum_{n=1}^{1000}i^n=0 \right ]$
v) $\frac{1}{1+i}=1*\frac{1-i}{(1+i)(1-i)} =\frac{1}{2} (1-i)$
vi) Let $z_1=x_1+iy_1 \: \: and \: \: z_2=x_2+iy_2$
$z_1+z_2=(x_1+x_2 )+i(y_1+y_2 )$
If $z_1+z_2$ is real then $y_1+y_2=0$
$y_1=-y_2$
$z_2=x_2-iy_1$
$z_2=x_1-iy_1 (\: \: when\: \: x_1=x_2 )$
So, $z_2=\bar{z_1}$
vii) $arg (z)+arg (\bar{z} )$
$If\: \: arg (z)= \theta , then \arg (\bar{z})=-\theta$
$\theta+(-\theta)=0$
viii) $|z+4|\leq 3$
$=|z+4-3|\leq|z+4|+|-3|$
$=|z+4-3|\leq3+3$
$=|z+4-3|\leq6$
The greatest value is 6 and the least value of $|z+1|$ is 0
ix)$\frac{z-2}{z+2}=\frac{\pi}{6}$
Let $z=x+iy$
$\left |\frac{x+iy-2}{x+iy+2} \right |=\left |\frac{(x-2)+iy}{(x+2)+iy} \right |= \frac{\pi}{6}$
$6|(x-2)+iy|=\pi|(x+2)+iy|$
$6\sqrt{(x-2)^2+y^2 }=\pi\sqrt{(x+2)^2+y^2 }$
$36[x^2+4-4x+y^2 ]=\pi ^2[x^2+4+4x+y^2 ]$
$36x^2+144-144x+36y^2=\pi^2 x^2+4\pi^2+4\pi^2 x+\pi^2 y^2$
$(36-\pi^2 ) x^2+(36-\pi^2 )-(144+4\pi^2 )x+144-4\pi^2=0$
Which represents an equation of a circle
x) $|z|=4$ and $arg (z) = \frac{5\pi}{6}$
Let $z=x+iy$
$|z|= \sqrt{x^2+y^2} =4$
$x^2+y^2=16$
$arg(z)=\tan^{-1}\frac{y}{x}=\frac{5\pi}{6}$
$\frac{y}{x}=\tan\left ( \pi-\frac{\pi}{6} \right )$
$=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$
$x=-\sqrt3 y$
$(-\sqrt3 y)^2+y^2=16$
$3y^2+y^2=16$
$4y^2=16$
$y^2=4$
$y=\pm2$
$x=-2\sqrt3$
$z=-2\sqrt3+2i$

Question 26 State True or False for the following:
i) The order relation is defined on the set of complex numbers.
ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction
iii) For any complex number z, the minimum value of |z| + |z – 11 is 1
iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1)
v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z2) = 0
vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z1 and Z2 be two complex numbers such that |z, + z2| = |z1 j + |z2|, then arg (z1 – z2) = 0.
(viii) 2 is not a complex number.

Answer:

(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex numbers can be compared. So, it is false.

(ii) Let $z= x+iy$

$z.i = (x+iy)i =xi= -y$ which rotates at angle of 180. So, it is ‘false’.

(iii) Let $z= x+iy$

$|z|+|z-1|= \sqrt{x^2+y^2} +\sqrt{(x-1)^2+y^2}$

The value of $|z|+|z-1|$ is minimum when $x=0,y=0 \, \: \: i.e.,1$

Hence, it is true.

iv) Let $z= x+iy$

Given that $\left | z-1 \right |=\left |z-i \right |$

$|x+iy-1|=|x+iy-i|$

$|(x-1)+iy|=|x-(1-y)i|$

$\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(1-y)}^2$

$(x-1)^2+y^2=x^2+(1-y)^2$

$x^2-2x+1+y^2=x^2+1+y^2-2y$

$-2x+2y=0 x-y=0$ which is a straight line slope=1

Now, the equation of the line through the point 1,0and 0,1

$y-0=\frac{1-0}{0-1} (x-1)$

$y=-x+1$ whose slope=-1

Multiplication of the slopes of two lines =-1*1=-1

So, they are perpendicular. Hence, it is true.

v)Let $z= x+iy$ $z\neq0 \: \: and \: \: Re(z)=0$

Since, the real part is 0 $x=0, z=0+iy=iy$

$1m(z^2 )=y^2 i^2=-y^2$which is real Hence, it is False.

vi) $(z-4)<|z-2|$

Let $z= x+iy$

$|x+iy-4|<|x+iy-2|$

$|(x-4)+iy|<|(x-2)+iy|$

$\sqrt{(x-4)^2+y^2 }<\sqrt{(x-2)^2+y^2}$

$(x-4)^2+y^2<(x-2)^2+y^2$

$(x-4)^2<(x-2)$

$x^2+16-8x<x^2+4-4x$

$8x+4x<-16+4$

$-4x<-12$

$x>3$ Hence, it is true.

vii) Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$

$|z_1+z_2 |=|z_1 |+|z_2 |$

$|x_1+iy_1+x_2+iy_2 |=|(x_1+iy_1 )+(x_2+y_2 i)|$

$= \sqrt{(x_1+x_2 )^2+(y_1+y_2 )^2 }$

$=\sqrt{x_1^2+y_1^2} +\sqrt{x_2^2+y_2^2 }$

Squaring both sides, we get $(x_1^2+x_2^2+2x_1 x_2+y_1^2+y_2^2+2y_1 y_2$)

$=x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

$=2x_1 x_2+2y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

$=x_1 x_2+y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

Again squaring on both sides we get $x_1^2 x_2^2+x_1^2 y_2^2+2x_1 y_1 x_2 y_2=x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 2x_1 y_1 x_2 y_2=x_1^2 y_2^2+x_2^2 y_1^2$

$x_1^2 y_1^2+x_2^2 y_1^2-2x_1 y_1 x_2 y_2=0$

$(x_1 y_2-x_2 y_1 )^2=0$

$x_1 y_2-x_2 y_1=0$

$x_1 y_2=x_2 y_1$

$\frac{x_1}{y_1}=\frac{x_2}{y_2}$

$\frac{y_1}{x_1}=\frac{y_2}{x_2}$

$arg(z_1)=arg(z_2)=0$ Hence, it is false.

(viii) Since every real number is a complex number. So, 2 is a complex number. Hence, it is false.

Question 27:

Answer:

a)Given $z=i+\sqrt3$

Polar form of z $=r[cos\theta+isin\theta]=i+\sqrt3$.

$r=\sqrt{3+1} =2$

And $\tan \alpha = \frac{1}{\sqrt3}$

$\alpha = \frac{\pi}{6}$

Since $x>0,y>0$

Polar form of z $=2[cos\frac{\pi}{6}+isin\frac{\pi}{6}]$

b) Given that $z=-1+\sqrt3=-1+\sqrt3 i$

Here argument z $=\tan^{-1}\left | \frac{\sqrt3}{-1} \right |= \tan^{-1}\sqrt{3}= \frac{\pi}{3}$

So, $\alpha = \frac{\pi}{3}$

Since, x<0 and y>0 $\theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

c) $|z+2|=|z-2|$

$|x+iy+2|=|x+iy-2|$

$|(x-2)+iy|=|(x-2)+iy|$

$\sqrt{(x+2)^2+y^2} =\sqrt{(x-2)^2+y^2}$

$(x+2)^2+y^2=(x-2)^2+y^2$

$(x+2)^2=(x-2)^2$

$x^2+4+4x=x^2+4-4x$

$8x=0 , x=0$

which represents the equation of the y-axis and is perpendicular to the line joining the points (-2,0) and (2,0)

d) $|z+2i|=|z-2i|$

$(x+iy+2i)=|x+iy-2i|$

$|x+(y+2)i|=|x+(y-2)i|$

$\sqrt{x^2+(y+2)^2} =\sqrt{x^2+(y-2)^2 }$

$x^2+(y+2)^2=x^2+(y-2)^2$

$(y+2)^2=(y-2)^2$

$y^2+4+4y=y^2+4-4y$

$8y=0 ,y=0$

which is the equation of the x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)

e) $|z+4i|\geq3$

$|x+iy+4|\geq3$

$|x+(y+4)i|\geq3$

$\sqrt{x^2+(y+4)^2} \geq3$

$x^2+(y+4)^2\geq9$

$x^2+y^2+8y+16 \geq9$

$x^2+y^2+8y+7 \geq0$ $r=\sqrt{(4)^2-7}=3$

which represents a circle on or outside having a centre (0,-4)

f) $|z+4|\leq3$

Let $z=x+iy$ Then, $|x+iy+4|\leq3$

$|(x+4)+iy|\leq3$

$\sqrt{(x+4)^2+y^2 } \leq3$

$x^2+8x+16+y^2\leq9$

$x^2+y^2+8x+7 \leq0$ which is a circle having centre (-4,0) and $r=\sqrt{(4)^2-7}=\sqrt9=3$ and is on the circle.

g) Let $z=\frac{1+2i}{1-i}$

$\frac{1+2i}{1-i}* \frac{1+i}{1+i}=\frac{1+i+2i+2i^2}{1-i^2}$

$=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}$

$=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}$

$=-\frac{1}{2}+\frac{3}{2}i$ which lies in third quadrant..

h) Given that z=1-i

Reciprocal of $z=\frac{1}{z}=\frac{1}{1-i}*\frac{1+i}{1+i}$

$=\frac{1+i}{1+i^2}$ which lies in first quadrant.

$So,(a)-(v),(b)-(iii),(c)-(i),(d)-(iv),(e)-(ii),(f)-(vi),(g)-(viii),(h)-(vii)$

Question 28 What is the conjugate of $\frac{2-i}{\left (1-2i \right )^2}$?

Answer:

$\frac{2-i}{\left (1-2i \right )^2}$$= \frac{2-i}{1+4i^2-4i}$
$= \frac{2-i}{-3-4i}= \frac{2-i}{-3-4i}*\frac{-3+4i}{-3+4i}$
$=\frac{-6+8i+3i-4i^2}{9-16i^2}$
$\frac{-6+11i+4}{9-16i^2}=\frac{-2+11i}{9+16}$
$\frac{-2+11i}{25}=\frac{-2}{25}+\frac{11}{25}i$
Hence, $\bar{z}=-\frac{2}{25}-\frac{11}{25}i$

Question 29 If $|z_1| = |z_2|$, is it necessary that $z_1 = z_2$?

Answer:

Let $z_1=x_1+iy_1 \: \: and\: \: z_2=x_2+iy_2$
$|x_1+iy_1 |=|x_2+iy_2 |$
$= \sqrt{x_1^2+y_1^2} = \sqrt{x_2^2+y_2^2}$
$x_1^2+y_1^2=x_2^2+y_2^2$
$\\$Consider $ z_1 = 3 + 4i \\ z_2 = 4 + 3i$
$z_1\neq z_2$ Hence, it is not necessary that $z_1= z_2$

Question 31 Find z if $|z| = 4$ and $arg (z) = 5\pi/6$..

Answer:

Polar form of $z=r[cos \theta+i\sin \theta]$
$=4\left [ \cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6} \right ]$
$=4\left [ \cos\left ( \pi-\frac{\pi}{6} \right )+i\sin \left ( \pi-\frac{\pi}{6} \right ) \right ]$
$=4\left [ -\cos\left (\frac{\pi}{6} \right )+i\sin \left (\frac{\pi}{6} \right ) \right ]$
$=4\left [-\frac{\sqrt3}{2}+i\frac{1}{2} \right ]$
$=-2\sqrt3+2i$

Question 32 Find $(1+i)\frac{(2+i)}{(3+i)}$.

Answer:

$\left | (1+i)\frac{(2+i)}{(3+i)}*\frac{3-i}{3-i} \right |=\left | (1+i).\frac{6-2i+3i-i^2}{9-i^2} \right |$
$=\left | \frac{(1+i)(7+i)}{9+1} \right |$
$=\left | \frac{7+i+7i+i^2}{9+1} \right |$
$=\left | \frac{7+i+7i+i^2}{10} \right |$
$=\left | \frac{7+8i-1}{10} \right |$
$=\left | \frac{6+8i}{10} \right |$
$=\left |\frac{3}{5}+\frac{4}{5}i \right |$
$=\sqrt{\left (\frac{3}{5} \right )^2+\left (\frac{4}{5} \right )^2}=1$

Question 33 Find the principal argument of $(1+i\sqrt3)^2$.

Answer:

$\begin{array}{c} (1+i \sqrt{3})^{2}=1+i^{2} \cdot 3+2 \sqrt{3} i \\ =1-3+2 \sqrt{3} i=-2+2 \sqrt{3} i \\ \tan \alpha=\left|\frac{2 \sqrt{3}}{-2}\right| \\ \tan \alpha=|-\sqrt{3}|=\sqrt{3} \\ \tan \alpha=\tan \frac{\pi}{3} \\ \alpha=\frac{\pi}{3} \end{array}$

Now Re(z)<0 and image(z)>0 $arg(z)=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$
$\text {Hence, the principal arg}=\frac{2 \pi}{3}$

Question 34 Where does z lie, if $|\frac{z-5i}{z+5i}|=1$.

Answer:

$\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{x+y i-5 i}{x+i y+5 i}\right|=1 \\ |x+(y-5) i|=|x+(y+5) i| \\ x^{2}+(y-5)^{2}=x^{2}+(y+5)^{2} \\ (y-5)^{2}=(y+5)^{2} \\ \qquad \begin{array}{r} y^{2}+25-10 y=y^{2}+25+10 y \\ 20 y=0 \quad y=0 \end{array} \end{array}$

Hence, z lies on the x-axis i.e., the real axis.

Question 35 $\sin x + i \cos 2x$ and $\cos x - i \sin 2x$ are conjugate to each other for: A. $x = n\pi$ B. $x = \left ( n+\frac{1}{2} \right )\frac{\pi}{2}$ C. $x = 0$ D. No value of x

Answer:

The answer is option (c).
let $z=\sin x+i \cos 2 x \quad \bar{z}=\sin x-i \cos 2 x$

But we are given that $\bar{z}=\cos x-i \sin 2 x$

$\sin x-i \cos 2 x=\cos x-i \sin 2 x$

Comparing the real and imaginary parts, we get

$\sin x=\cos x \quad$ and $\quad \cos 2 x=\sin 2 x$

$\tan \mathrm{x}=1 \text { and } \tan 2 \mathrm{x}=1$
$\tan 2 \mathrm{x}= \frac{2\tan x }{1-\tan^2 x}=1$
The above value is not satisfied by tan x = 1.
Hence, no value of x is possible.

Question 36 The real value of α for which the expression $\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}$ is purely real is: A.$\left ( n+1 \right )\frac{\pi}{2}$ B.$\left ( 2n+1 \right )\frac{\pi}{2}$ C.$n\pi$ D. None of these

Answer:

The answer is the option (c).
Let $z=\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}$
$\begin{aligned} &\begin{array}{c} =\frac{(1-i\sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)} \\\\ =\frac{1-2 i \sin \alpha-i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{(1)^{2}-(2 i \sin \alpha)^{2}} \\\\ =\frac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}=\frac{\left(1-2 \sin ^{2} \alpha\right)-3 i \sin \alpha}{1+4 \sin ^{2} \alpha} \\\\ \frac{-3 i \sin \alpha}{1+4 \sin ^{2} \alpha}=0 \\\\ \sin \alpha=0 \quad \operatorname{} \alpha=n \pi \end{array}\\ &\text { Hence, c is correct. } \end{aligned}$

Question 37 If z = x + iy lies in the third quadrant, the $\frac{\bar{z}}{z}$also lies in the third quadrant if A. x > y > 0 B. x < y < 0 C. y < x < 0 D. y > x > 0

Answer:

The answer is the option (b).
If z lies in the third quadrant,So,$x<0 \: \: and\: \: y<0\: \: , z=x+iy$
$\begin{aligned} \frac{\bar{z}}{z} &=\frac{x-i y}{x+i y}=\frac{x-i y}{x+i y} * \frac{x-i y}{x-i y} \\ &=\frac{x^{2}+i^{2} y^{2}-2 x y i}{x^{2}-i^{2} y^{2}} \\ &=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}-\frac{2 x y}{x^{2}+y^{2}} i \end{aligned}$
When z lies in the third quadrant then $\frac{\bar{z}}{z}$ will also be in the third quadrant.
$\begin{array}{c} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \text { and } \frac{2 x y}{x^{2}+y^{2}}>0 \\\\ x^{2}-y^{2}<0 \text { and } 2 x y>0 \\\\ x^{2}<y^{2} \text { and } x y>0 \\\\ \text { So, } x<y<0 \end{array}$
Hence, b is correct.

Question 38 The value of $(z + 3) (\bar{z}+3)$ is equivalent to A.$|z + 3|^2$ B. |z – 3| C.$z^2 + 3$ D. None of these

Answer:

The answer is the option (a).
$Let\: \: z=x+iy \: \: So,\: \: (z+3)(\bar{z} +3)=(x+iy+3)(x-iy+3)$
$=\left [ \left ( x+3 \right ) +iy\right ]\left [ \left ( x+3 \right ) -iy\right ]$
$=(x+3)^2-y^2 i^2=(x+3)^2+y^2$
$=|x+3+iy|^2=|z+3|^2$

Question 39 If $\left ( \frac{1+i}{1-i} \right )^{x}=1$ , then A. $x = 2n + 1$ B. $x = 4n$ C. $x = 2n$ D.$x = 4n + 1$

Answer:

The answer is the option(b).
$\begin{array}{c} \left(\frac{1+i}{1-i}\right)^{x}=1 \\\\ \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{x}=1 \\\\ \left(\frac{1-1+2 i}{1+1}\right)^{x}=1 \\\\ \left(\frac{2 i}{2}\right)^{x}=1(i)^{x}=1 \end{array}$
$x = 4 n, n \epsilon N$
Hence, the correct option is

Question 40: A real value of x satisfies the equation $\frac{3-4 i x}{3+4 i x}=\alpha-i \beta$ ($\alpha, \beta \in R$) , if $\alpha ^2 +\beta ^2 =$ A. 1 B. –1 C. 2 D. –2

Answer:

The answer is the option (a).

$\begin{aligned} &\text {Given that} \frac{3-4 i x}{3+4 i x}=\alpha-i \beta\\ &=\frac{3-4 i x}{3+4 i x} * \frac{3-4 i x}{3-4 i x}=\alpha-i \beta\\ &\frac{9-12 i x-12 i x+16 i^{2} x^{2}}{9-16 i^{2} x^{2}}=\alpha-i \beta\\ &=\frac{9-16 x^{2}}{9+16 x^{2}}-\frac{24 x}{9+16 x^{2}}=\alpha-i \beta \ldots \ldots .(i)\\ &\frac{9-16 x^{2}}{9+16 x^{2}}+\frac{24 x}{9+16 x^{2}}=\alpha+i \beta \ldots \ldots \ldots(ii)\\\end{aligned}$
Multiplying eqn (i )and (ii )we get
$\left (\frac{9-16 x^{2}}{9+16 x^{2}} \right )^{2}+\left (\frac{24 x}{9+16 x^{2}} \right )^{2}=\alpha^{2}+ \beta^{2}$
$\frac{81+256 x^{4}-288 x^{2}+576x^2}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}$
$\begin{aligned}\frac{81+256 x^{4}+288 x^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}\\ \frac{\left(9+16 x^{2}\right)^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}=1 \end{aligned}$

Question 41 Which of the following is correct for any two complex numbers $z_1\: \: and\: \: \: \: z_2$? A. $|z_1 z_2| = |z_1||z_2|$ B. $arg (z_1z_2) = arg (z_1). Arg (z_2)$ C. $|z_1 + z_2| = |z_1|+ |z_2|$ D. $|z_1 + z_2| \geq |z_1| - |z_2|$

Answer:

The answer is the option (a).
$\begin{array}{c} z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{2}\right) \\ \left|z_{1}\right|=r_{1} \\ z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \left|z_{2}\right|=r_{2} \\\\ z_1z_2=r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right] \\ =r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right) \cdot r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \end{array}$
$\begin{array}{c} =r_{1} r_{2}\left[\left(\cos \theta_{1} \cos \theta_{2}-\sin \theta_{1} \sin \theta_{2}\right)+i\left(\sin \theta_{1} \cos \theta_{2}+\cos \theta_{1} \sin \theta_{2}\right)\right] \\ =r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right]\left|z_{1} z_{2}\right|=\left|z_{1} \| z_{2}\right| \end{array}$

Hence, option A is correct.

Question 42 The point represented by the complex number $2 -i$ is rotated about the origin through an angle $\pi/2$ in the clockwise direction, the new position of a point is: A. $1 + 2i$ B. $-1-2i$ C. $2 + i$ D. $-1 + 2i$

Answer:

The answer is the option (b).
If z rotated through an angle of $\pi/2$ about the origin in a clockwise direction.
Then the new position $=z.e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).\left [ \cos\left (- \frac{\pi}{2} \right )+isin\left (- \frac{\pi}{2} \right ) \right ]$
$=\left (2-i \right ).\left (0-i \right )$
$=-1-2i$
Hence, the correct option is (b)

Question 43 Let $x, y \in R$, then $x + iy$ is a non-real complex number if: A. $x = 0$ B.$y = 0$ C. $x \neq 0$ D. $y \neq 0$

Answer:

The answer is the option (d).
$x+iy$ is a non-real complex number if $y\neq0$,

Hence, d is correct

Question 44 If $a + ib = c + id$, then A. $a^2 + c^2 = 0$ B. $b^2 + c^2 = 0$ C. $b^2 + d^2 = 0$ D. $a^2 + b^2 = c^2 + d^2$

Answer:

The answer is the option (d).
$a+ib=c+id$
$|a+ib|=|c+id|$
$\sqrt{a^2+b^2} =\sqrt{c^2+d^2}$
Squaring both sides, $a^2+b^2=c^2+d^2$
Hence, d is correct

Question 45 The complex number z which satisfies the condition $\left |\frac{i+z}{i-z} \right |=1$ lies on A. circle $x^2 + y^2 = 1$ B. the x-axis C. the y-axis D. the line $x + y = 1$

Answer:

The answer is the option (b).
$\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{i+x+i y}{i-x-y i}\right|=1 \\\\ \left|\frac{x-(y+1) i}{-x-(y-1) i}\right|=1 \\\\ |x+(y+1) i|=|-x-(y-1) i| \\\\ \sqrt{x^{2}+(y+1)^{2}}=\sqrt{x^{2}+(y-1)^{2}} \\\\ x^{2}+(y+1)^{2}=x^{2}+(y-1)^{2} \\\\ y^{2}+2 y+1=y^{2}-2 y+1 \\ 2 y=-2 y \\\\ y=0 ; x-a x i s \end{array}$

Hence, b is correct

Question 46 If z is a complex number, then A.$|z^2| > |z|^2$ B. $|z^2| \: \: equals\: \: |z|^2$ C.$|z^2| < |z|^2$ D. $|z^2| \geq |z|^2$

Answer:

. The answer is the option (b).
$Let \: \: z=x+iy \: \: \: \: ,|z|^2=x^2+y^2$
$Now,z^2=x^2+y^2 i^2+2xyi$
$z^2=x^2-y^2+2xyi$
$|z|^2=\sqrt{(x^2-y^2 )^2+(2xy)^2 }$
$=\sqrt{x^4+y^4+2x^2 y^2} =\sqrt{(x^2+y^2 )^2}$
$|z|^2=x^2+y^2=|z|^2$

Hence, b is correct

Question 47 $|z_1 + z_2| = |z_1| + |z_2|$ is possible if A. $z_1 =\bar{z_1}$ B. z2=z1 C. $arg (z_1) = arg (z_2)$ D. $|z_1| = |z_2|$

Answer:

The answer is the option (c).
Since, $|z_1+z_2 |=|z_1 |+|z_2 |$
$|z_1+z_2 |=|r_1 \left ( \cos \theta _1+i\sin\theta _1 \right )+r_2 \left ( \cos \theta _2+i\sin\theta _2 \right )|$
$=\sqrt{r_{1}^{2} \cos ^{2} \theta_{1}+r_{2}^{2} \cos ^{2} \theta_{2}+2 r_{1} r_{2} \cos \theta_{1} \cos \theta_{2}+r_{1}^{2} \sin ^{2} \theta_{1}+r_{2}^{2} \sin ^{2} \theta_{2}+2 r_{1} r_{2} \sin \theta_{1} \sin \theta_{2}}$
So, $\begin{array}{c} =\sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)} \\ \left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right| \\\\ \sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}=r_{1}+r_{2} \end{array}$
Squaring both sides,
$\begin{array}{c} r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \\ 2 r_{1} r_{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=0 \\ 1-\cos \left(\theta_{1}-\theta_{2}\right)=0 \\ \cos \left(\theta_{1}-\theta_{2}\right)=1 \\ \theta_{1}-\theta_{2}=0 \\ \theta_{1}=\theta_{2} \\ \operatorname{so} \arg \left(z_{1}\right)=\arg \left(z_{2}\right) \end{array}$

Question 48 The real value of θ for which the expression $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is a real number is: A. $n\pi+\frac{\pi}{4}$ B. $n\pi+(-1)^n\frac{\pi}{4}$ C. $2n\pi+\frac{\pi}{2}$ D. none of these

Answer:

Let $z=\frac{1+i \cos \theta}{1-2 i \cos \theta}=\frac{1+i \cos \theta}{1-2 i \cos \theta} * \frac{1+2 i \cos \theta}{1+2 i \cos \theta}$

On solving, we get
$\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1-4 i^{2} \cos ^{2} \theta} \\=\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1+4 \cos ^{2} \theta}$
$=\frac{\left(1-2 \cos ^{2} \theta\right)+3 i \cos \theta}{1+4 \cos ^{2} \theta}$
If z is a real number, then
$\frac{3 \cos \theta}{1+4 \cos ^{2} \theta}=0$
$\\3 \cos \theta=0 \\\cos \theta=0$
$\theta=\frac{(2 n+1) \pi}{2}, n \in N$

Question 49 The value of $arg (x)$ when $x < 0$is: A. 0 B. $\frac{\pi}{2}$ C. π D. none of these

Answer:

The answer is option (c).
$\\ Let\: \: z= -x+0i \: \: and \: \: x<0 \\ |z|=\sqrt{(-1)^2+(0)^2 }=1,x<0$

Since, the point (-x,0)lies on the negative side of the real axis, principal argument (z)=π

Hence, option c is correct.

Question 50 If $f(z)=\frac{7-z}{1-z^2}$where $z = 1 + 2i$, then $|f(z)|$ is A. $\frac{\left | z \right |}{2}$ B. $|z|$ C. $2|z|$ D. none of these

Answer:

The answer is option (a).
Given that $\\ z=1+2i \\ |z|=\sqrt{(1)^2+(2)^2 }=\sqrt5$
Now $f(z)=\frac{7-z}{1-z^2}$
$\begin{aligned} =\frac{7-(1+2 i)}{1-(1+2 i)^{2}} &=\frac{7-1-2 i}{1-1-4 i^{2}-4 i} \\ =\frac{6-2 i}{4-4 i} &=\frac{3-i}{2-2 i} \end{aligned}$
$\begin{aligned} =\frac{3-i}{2-2 i} * \frac{2+2 i}{2+2 i}=\frac{6+6 i-2 i-2 i^{2}}{4-4 i} \\ =\frac{6+4 i+2}{4+4} &=\frac{8+4 i}{8} \\ =1+\frac{1}{2} i & \end{aligned}$
$f(z)=\sqrt{(1)^{2}+\left(\frac{1}{2}\right)^{2}}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}$

Hence, option A is correct.

NCERT Solutions for Class 11 Mathematics Chapter Wise

Careers360 has put all NCERT Class 11 Maths Solutions in one place to help students. Use the links below to open them.

Important Notes from NCERT Exemplar Class 11 Maths Solution Chapter 5

Various laws need to be followed and derived by a mathematician when it comes to multiplying two complex numbers. All of them are named after the features of the specific formula derivation. Ahead, you will also learn about the Modulus and Conjugate of complex numbers and how to solve sums with their help. There is an Argand and Polar representation in the Chapter, relating to graphical presentations. A point needs to be derived at the ‘x’ and ‘y’ axes to form an Argand or complex plane.

Students, with the help of NCERT Exemplar Class 11 Maths Solutions Chapter 5, will not face any issues when trying to solve such problems. It will also help them score better in exams.

All the concepts have been covered in the NCERT Exemplar Solutions for Class 11 Maths Chapter 5. By using the NCERT Exemplar Class 11 Maths Chapter 5 Solutions PDF Download function, students can access quality study material that is effectively constructed by experts for the best learning experience.

NCERT Solutions for Class 11 Mathematics Chapterwise

Careers360 has gathered the entire set of NCERT Class 11 Maths Exemplar Solutions in a single spot for student convenience. Access them through the links given below.

Importance of Solving NCERT Questions for Class 11 Chapter 5 Complex Numbers and Quadratic Equations

Complex numbers and quadratic equations are not only useful in Class 11 but also for higher studies and competitive exams. Strengthening basic concepts is a necessity for students so that later they do not face any difficulties solving Complex numbers and quadratic equations questions in higher studies or competitive examinations.
Some important facts about solving Complex numbers and quadratic equations in Class 11 are listed below.

• Students can study strategically at their own pace after accessing Class 11 Maths NCERT Solutions Chapter 5. This will boost their confidence to attempt other questions from this Chapter.

• Class 11 Maths Chapter 5 NCERT Solutions are solved by subject-matter experts and are very reliable at the same time. The Solutions provide shortcuts as well as detailed explanations with necessary formulae that will help students to understand the answers better.

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• NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations is designed to give the students step-by-step Solutions for a particular question.

NCERT Solutions of Class 11 - Subject-wise

Here are the subject-wise links for the NCERT Solutions of Class 11:

• NCERT Solutions for Class 11 Maths
• NCERT Solutions for Class 11 Physics
• NCERT Solutions for Class 11 Chemistry
• NCERT Solutions for Class 11 Biology

NCERT Notes of Class 11 - Subject Wise

Given below are the subject-wise NCERT Notes of Class 11 :

• NCERT Notes for Class 11 Maths
• NCERT Notes for Class 11 Physics
• NCERT Notes for Class 11 Chemistry
• NCERT Notes for Class 11 Biology

NCERT Books and NCERT Syllabus

Before a new academic session begins, students should check the latest syllabus to know which chapters are included. Below are the updated syllabus links along with suggested reference books.

• NCERT Books Class 11 Maths
• NCERT Syllabus Class 11 Maths
• NCERT Books Class 11
• NCERT Syllabus Class 11

NCERT Exemplar Class 11 Solutions

Given below are the subject-wise Exemplar solutions of Class 11 NCERT:

• NCERT Exemplar Class 11 Biology Solutions
• NCERT Exemplar Class 11 Chemistry Solutions
• NCERT Exemplar Class 11 Physics Solutions